Let P be what A becomes when h = 0, then P is a function of r alone, and reasoning as above we prove and proceeding as before to separate the parts of this new function of z, we find Then we have B=Q+ Ch f(x + h) = f(x) + Ah AP+Bh f(x + h) = f(x) + P. h + B¿1 BQ+ Ch f(x+h) = f(x) + P.h+Q.h2 + Ch3 when P, Q are functions of x alone, and proceeding in this manner, we get f(x+h) developed in the required form. PROP, If the variable of a function be supposed to consist of two parts, the diffe rential co-efficient will be the same to whichever part the variation be ascribed. Then it is required to prove that the differential co-ethicient will be the same when we consider ≈ variable and h constant, as it will be when we con sider h variable and a constant. 1. If we consider x variable, then by art. 4. сар. III. Having fully proved that if y = f (x), and that if x become r+h, so long as a remains indeterminate we shall have f(x+h) = f (x) + Ph + Q42 + Rh3 + ... where P, Q, R, . . . are functions of x and do not invoive h (1) Let us take the first differential co-efficient of ƒ (x + h) upon the supposition that x is constant and h variable d f (x + h) dh =P + 2Qh + 3Rh2 + (2) Again, take the first differential co-efficient of ƒ (x + h) upon the supposition that a is variable and h constant comparing therefore the homologous terms in the identical series (2) and (3) Ahich is the series known by the name of TAYLOR'S THEOREM, perhaps the most important in the whole range of pure mathematics. dy dy dy Sometimes for the sake of brevity the differential co-efficients dr'd dr are represented by p, q, r, • • • respectively, in which case the series may be written h2 h3 + ... y' or f (x + h) = y +ph+q· 1.2 +r. According to the notation of Lagrange, the first differential co-efficient of f(x), or as he designates it, the first derived function of ƒ (x) is represented by f'(x), the second differential co-efficient by f" (x), the third by ƒ” (x), and so on, in his works, therefore, the above series appears under the form h f(x + h) = f (x)+ƒ' (x) ï+ƒ" (x) i ̈. z+ƒ” (x) h for+h, the series will become dy h d3y h3 2.3 + in which the terms are alternately positive and negative. TAYLOR'S THEOREM AND MACLAURIN'S THEOREM. Let y be a function of x, which it is possible to develop in a series of poɛ tive ascending powers of that variable, and let us suppose that y= A + Bx + Cx2 + Dx3 + Ex1 + .... .... and when a becomes x+h, let y becoms y'; then we have (1) y' = A + B(x + h) + C(x + h)2 + D (x + h)3 +.... +Bh+2Cxh + 3Dx2h + + Ch2+3Dxh2 + + Dh3 + (2) But by eq. (1) we have the first, second, &c. differential co-efficients, as follows: h2 hence by multiplying these differential co-efficients respectively by h, 1.2 13 1.2.3' y = y + dy h+d2y h2 + &c. and substituting the results in equation (2) we have finally d3y h3 Again, since the co-efficients A, B, C, D, &c. in (1) do not involve x, they will remain unchanged whatever value be assigned to x. Let then the particular values of y, and its successive differential co-efficients, be expressed by means of brackets, and when x=0, we shall have by (1) and (3) +..... (A) The former of these equations (A) is Taylor's theorem, and the latter (B) is Maclaurin's theorem; and the demonstrations we have given of these most important theorems will be readily comprehended by the student. We regret that room will not permit us to exemplify the latter of these theorems. 2.3 (F), &c. (B) Cases in which Taylor's Theorem fails. In the preceding demonstration of Taylor's theorem, we have supposed with regard to ƒ (x) that a remains indeterminate, and .. that ƒ (x) has as many values as f(x+1). But when we assign particular values to ≈, the above reasonings will not always hold good, and we shall not in all cases obtain the true expansion of ƒ (x + h) . . . . .... 1. Let f (x) = x when x = a, then the expansion of ƒ (≈ + h) must contain negative powers of h. For a will be determined from the equation n being any positive whole number whatever, and ◊ (x) some function of z which does not become 0 or a when x = a 2. Let f (x) contain a radical which disappears when x = a. In this case, either the radical itself must vanish when ra, or its co-effi cient must vanish. m If the radical itself vanish in ƒ (x) when = a, it must be of the form (z mand n being whole numbers; hence f(x+h) will contain the corresponding radical a + h) which, on making x = a becomes hä' so that the developement of ƒ (a + h) according to powers of h may contain the radical hand its powers. (x If the co-efficients of the radical vanish when x = a, then this co-efficient must be of the form (x — a)a, ʼn being a whole number, in this case the radical will disappear in the differential co-efficients ƒ' (a), ƒ" (a) . . . . ƒa—1 (@), but will be found in those of higher orders. In general the following propo sition will hold good: When we assign a particular value to x in the developement of ƒ (x + h), if a term appear containing a fractional power of h which lies between h and h"+', then Taylor's theorem will hold good for the first n terms only. Let +N++++.. (a + h) = A + Bh + Ch3 + Dh3 + . . . + M/" + Nh® +++L+' + ... Take the differential co-efficients regarding has the variable, and let us denote the successive co-efficients by f'(a + h), ƒ“(a + k), &c. f'(a+4)= B+2Ch + 3Dh2 + ... + M'h11 + Nh" + −1+L1⁄2" + ... h) f"(a + h) = 1.2.C + 2.3.Dh + ... + MTM1⁄2~2 + No1⁄2»+÷¬2+L1⁄2a1+.. .... The co-efficients A, B. C, . . . . are the values which ƒ (x) and its differential co-efficients assume when x = a, precisely as in the series of Taylor. But at each differentiation the first term disappears because it is constant. When we arrive at the nth differential co-efficient, on the supposition that h = 0, and all the succeeding differential co-efficients will, in like manner, be infinite. It only now remains for us to show how we can obtain the developement of f(x+h) when Taylor's theorem fails, If, then, we wish to obtain the developement of f (x + h) when x = ɑ, we must calculate the terms of the series but if, in effecting this calculation, we find that one of the differential coefficients becomes infinite upon the supposition that xa, we must employ the following process. Substitute (x + h) for x in f (x); then the term which contains x -a in the denominator, will now contain x -ah, and will no longer become infinite u hen xa, but will become a term involving a fractional power of h. ƒ (x + h) = 2ax − x2 + a √ œa − x2 + { 2 (a − 1) + But, when x = a the term multiplied by h becomes infinite, hence Taylor theorem fails and the developement is no longer possible. ах |